Modular Arithmetic Practice Questions — JAMB UTME

Find the smallest non-negative integer $x$ that satisfies the linear modular congruence $3x \equiv 8 \pmod{11}$ .

Show answer & explanation

Answer: 10

Answer

The smallest non-negative integer

x

is 10.

Evaluating

3x \equiv 8 \pmod{11}

by testing multiples of

11

added to

8

gives

30

, which divided by

3

yields

x = 10

. Since

10 \in [0, 10]

, it is the canonical solution.

Step-by-Step Solution

1

Convert the modular congruence into an algebraic equation

3x = 8 + 11k

for an integer

k

By definition of congruence modulo

11

,

3x - 8

must be a multiple of

11

.

2

Find the smallest integer

k \ge 0

such that

8 + 11k

is divisible by 3

When

k = 2

,

8 + 11(2) = 30

30

is divisible by

3

(

30 / 3 = 10

).

3

Divide by 3 to isolate

x

x = 10

3(10) = 30 \equiv 8 \pmod{11}

.

Key Concept

Linear Modular Congruence

Estimated Time:1m 15s

Question 2Question →

What is the value of $(-38) \pmod{7}$ expressed in standard non-negative remainder form?

4

-3

3

5

Show answer & explanation

Answer: 4

Answer

4

The value

4

is correct because

-38 = 7 \times (-6) + 4

. The remainder

4

lies in the standard non-negative range

0 \le r < 7

. Alternatively, adding multiples of

7

to

-38

gives

-38 + 35 = -3

, and adding

7

once more gives

-3 + 7 = 4

.

Step-by-Step Solution

1

Find the largest multiple of the modulus

7

that is less than or equal to

-38

.

The multiple is

7 \times (-6) = -42

.

Modular arithmetic requires the remainder

r

to satisfy

0 \le r < 7

in standard form.

2

Calculate the remainder by subtracting the multiple from the dividend.

-38 - (-42) = -38 + 42 = 4

.

The remainder is the non-negative difference between the number and the multiple of the modulus.

Key Concept

Modular Arithmetic with Negative Numbers

Estimated Time:1m 0s

Question 3Question →

If $5x \equiv 2 \pmod{11}$ , what is the value of $(x^3 - 4x) \pmod{11}$ expressed in standard non-negative remainder form?

7

-4

4

8

Show answer & explanation

Answer: 7

Answer

7

Solving the linear congruence

5x \equiv 2 \pmod{11}

gives

x \equiv 7 \pmod{11}

. Substituting

x = 7

into the expression

(x^3 - 4x)

yields

7^3 - 4(7) = 343 - 28 = 315

. Dividing 315 by 11 gives a quotient of 28 and a remainder of 7. Alternatively, working entirely modulo 11 gives

2 - 6 = -4 \equiv 7 \pmod{11}

. Thus, the canonical non-negative remainder is 7.

Step-by-Step Solution

1

Find the modular inverse of 5 modulo 11

The inverse is 9, since

5 \times 9 = 45 \equiv 1 \pmod{11}

.

To isolate

x

in the linear congruence

5x \equiv 2 \pmod{11}

, multiply both sides by the modular inverse of 5.

2

Solve for x modulo 11

x \equiv 2 \times 9 = 18 \equiv 7 \pmod{11}

.

Simplifying 18 modulo 11 gives the canonical value of

x

.

3

Evaluate

x^3 - 4x

modulo 11

x^3 \equiv 7^3 = 343 \equiv 2 \pmod{11}

and

4x \equiv 4(7) = 28 \equiv 6 \pmod{11}

, so

x^3 - 4x \equiv 2 - 6 = -4 \pmod{11}

.

Substitute

x \equiv 7

into the polynomial expression and reduce each term modulo 11.

4

Convert negative remainder to canonical non-negative form

-4 + 11 = 7 \pmod{11}

.

Modular remainders must be expressed within the standard range

[0, 10]

.

Key Concept

Linear modular congruences and negative remainder canonical reduction

Estimated Time:2m 0s

Question 4Question →

What is the smallest non-negative integer $k$ that satisfies the modular congruence $7^{99} + k \equiv -15 \pmod{11}$ ?

10

1

7

8

Show answer & explanation

Answer: 10

Answer

The correct answer is 10.

Using Fermat's Little Theorem,

7^{10} \equiv 1 \pmod{11}

, which simplifies

7^{99} \pmod{11}

to

7^9 \equiv 8 \pmod{11}

. Reducing the right-hand side gives

-15 \equiv 7 \pmod{11}

. The modular equation

8 + k \equiv 7 \pmod{11}

yields

k \equiv -1 \pmod{11}

. Adding the modulus

11

gives the canonical positive remainder

10

.

Step-by-Step Solution

1

Simplify

7^{99} \pmod{11}

using Fermat's Little Theorem.

7^{10} \equiv 1 \pmod{11}

, so

7^{99} = (7^{10})^9 \times 7^9 \equiv 1^9 \times 7^9 \equiv 7^9 \pmod{11}

.

Since

11

is prime and

\gcd(7, 11) = 1

, Fermat's Little Theorem allows exponent reduction modulo

10

.

2

Compute

7^9 \pmod{11}

.

7^1 \equiv 7

,

7^2 = 49 \equiv 5

,

7^4 \equiv 5^2 = 25 \equiv 3

,

7^5 \equiv 7 \times 3 = 21 \equiv 10

,

7^9 = 7^5 \times 7^4 \equiv 10 \times 3 = 30 \equiv 8 \pmod{11}

.

Repeated squaring and modular multiplication efficiently reduces

7^9

modulo

11

.

3

Reduce the right-hand side

-15 \pmod{11}

.

-15 = -2(11) + 7 \equiv 7 \pmod{11}

.

Converting negative numbers into the standard non-negative remainder range

[0, 10]

.

4

Substitute remainders into the congruence and solve for

k

.

8 + k \equiv 7 \implies k \equiv 7 - 8 = -1 \pmod{11}

.

Linear algebraic rearrangement in modular arithmetic.

5

Convert the negative remainder

-1

to canonical non-negative form.

k = -1 + 11 = 10

.

The standard remainder must satisfy

0 \leq k < 11

.

Key Concept

Modular Exponentiation & Negative Remainder Reduction

Question 5Question →

Find the smallest positive integer $n$ that simultaneously satisfies the linear modular congruences $3n \equiv 5 \pmod{13}$ and $4n \equiv 2 \pmod{9}$ .

Show answer & explanation

Answer: 32

Answer

The smallest positive integer satisfying both congruences is 32.

Solving the first congruence

3n \equiv 5 \pmod{13}

gives

n \equiv 6 \pmod{13}

(since

3 \times 9 \equiv 1 \pmod{13}

and

5 \times 9 = 45 \equiv 6 \pmod{13}

), which means

n

can be written as

13k + 6

. Substituting this into the second congruence

4n \equiv 2 \pmod{9}

yields

4(13k + 6) \equiv 2 \pmod{9} \implies 52k + 24 \equiv 2 \pmod{9}

. Reducing the coefficients modulo 9 gives

7k + 6 \equiv 2 \pmod{9} \implies 7k \equiv -4 \equiv 5 \pmod{9}

. Multiplying by 4 (the modular inverse of 7 modulo 9) yields

k \equiv 20 \equiv 2 \pmod{9}

. Setting

k = 2

yields the smallest positive integer

n = 13(2) + 6 = 32

.

Step-by-Step Solution

1

Solve the first modular congruence

3n \equiv 5 \pmod{13}

for

n

n \equiv 6 \pmod{13}

, which implies

n = 13k + 6

Multiplying both sides by the modular inverse of 3 modulo 13 (which is 9) isolates

n

.

2

Solve the second modular congruence

4n \equiv 2 \pmod{9}

for

n

n \equiv 5 \pmod{9}

Multiplying both sides by the modular inverse of 4 modulo 9 (which is 7) isolates

n

.

3

Substitute

n = 13k + 6

into

n \equiv 5 \pmod{9}

and simplify modulo 9

4k \equiv 8 \pmod{9}

Reducing 13 modulo 9 yields

4k

, and subtracting 6 from 5 yields

-1 \equiv 8 \pmod{9}

.

4

Solve for

k

and calculate the smallest positive integer

n

k \equiv 2 \pmod{9}

, giving

n = 13(2) + 6 = 32

Setting the integer parameter

k

to its minimum non-negative value

2

provides the smallest positive integer solution.

Key Concept

System of Linear Modular Congruences and Modular Inverses

Question 6Question →

If $x$ is the smallest positive integer satisfying the modular congruence $2^x \equiv 7 \pmod{11}$ , what is the value of $(3x^2 - 4x + 5) \pmod{11}$ expressed in standard non-negative remainder form?

Show answer & explanation

Answer: 3

Answer

The smallest positive integer exponent satisfying

2^x \equiv 7 \pmod{11}

is

x = 7

. Substituting

x = 7

into

3x^2 - 4x + 5

yields

124

, which simplifies to

3 \pmod{11}

.

Evaluating powers of 2 modulo 11 shows that

2^7 = 128 \equiv 7 \pmod{11}

, giving

x = 7

. Substituting

x = 7

into

3x^2 - 4x + 5

gives

124

, which leaves a remainder of

3

when divided by

11

.

Step-by-Step Solution

1

Find the smallest positive integer exponent

x

satisfying

2^x \equiv 7 \pmod{11}

x = 7

Evaluating consecutive powers of 2 modulo 11 shows

2^1 \equiv 2

,

2^2 \equiv 4

,

2^3 \equiv 8

,

2^4 \equiv 5

,

2^5 \equiv 10

,

2^6 \equiv 9

, and

2^7 \equiv 7

, making

x = 7

the smallest positive integer power.

2

Substitute

x = 7

into the expression

3x^2 - 4x + 5

124

Direct substitution gives

3(7)^2 - 4(7) + 5 = 3(49) - 28 + 5 = 147 - 28 + 5 = 124

.

3

Reduce 124 modulo 11 to standard non-negative remainder form

3

Dividing 124 by 11 yields a quotient of 11 with a remainder of 3 (

124 = 11 \times 11 + 3

).

Key Concept

Modular Exponentiation and Algebraic Evaluation in Modular Arithmetic

Question 7Question →

What is the value of $(17 - 42) \pmod{9}$ expressed in standard non-negative remainder form?

7

2

5

4

Show answer & explanation

Answer: 2

Answer

The value of

(17 - 42) \pmod{9}

in standard non-negative remainder form is

2

.

Evaluating

17 - 42

gives

-25

. Reducing

-25

modulo

9

requires finding a non-negative remainder

r

such that

-25 = 9q + r

with

0 \le r < 9

. Choosing

q = -3

yields

-25 = 9(-3) + 2

, making the correct non-negative remainder

2

. Alternatively,

17 \equiv 8 \pmod{9}

and

42 \equiv 6 \pmod{9}

, so

8 - 6 = 2 \pmod{9}

.

Step-by-Step Solution

1

Perform the subtraction within the parentheses.

17 - 42 = -25

Evaluate the arithmetic expression inside the modulo operation first.

2

Express

-25

in terms of the modulus

9

using the division algorithm

a = qn + r

, where

0 \le r < n

.

-25 = 9 \times (-3) + 2

The remainder

r

must be non-negative (

0 \le r < 9

).

3

Identify the canonical non-negative remainder.

r = 2

Alternatively,

-25 \equiv -7 \pmod{9}

, and adding the modulus gives

-7 + 9 = 2

.

Key Concept

Modular Arithmetic and Negative Remainder Reduction

Estimated Time:45s

Question 8Question →

A binary operation $\Delta$ defined on the set of integers modulo $11$ is given by $a \Delta b \equiv (2a^2 - 3ab + b^2) \pmod{11}$ . Find the smallest non-negative integer $x$ that satisfies the equation $4 \Delta x \equiv 5 \pmod{11}$ .

Show answer & explanation

Answer: 3

Answer

The smallest non-negative integer

x

that satisfies

4 \Delta x \equiv 5 \pmod{11}

is 3.

Evaluating

4 \Delta x

gives

32 - 12x + x^2 \equiv x^2 - x + 10 \pmod{11}

. Setting this congruent to

5 \pmod{11}

yields

x^2 - x + 5 \equiv 0 \pmod{11}

, which converts to

x^2 - x - 6 \equiv 0 \pmod{11}

. Factoring gives

(x - 3)(x + 2) \equiv 0 \pmod{11}

, which yields solutions

x \equiv 3 \pmod{11}

and

x \equiv 9 \pmod{11}

. The smallest non-negative integer among these solutions is

3

.

Step-by-Step Solution

1

Substitute the given value

a = 4

into the operation definition.

4 \Delta x = 2(4)^2 - 3(4)x + x^2 = 32 - 12x + x^2

This establishes the explicit algebraic polynomial in terms of

x

.

2

Reduce coefficients modulo

11

.

32 \equiv 10 \pmod{11}

and

-12x \equiv -x \pmod{11}

, giving

x^2 - x + 10 \pmod{11}

.

Simplifying coefficients reduces computational complexity during equation solving.

3

Form the modular quadratic equation and set it to zero.

x^2 - x + 10 \equiv 5 \pmod{11} \implies x^2 - x + 5 \equiv 0 \pmod{11} \implies x^2 - x - 6 \equiv 0 \pmod{11}

.

Expressing

5 \equiv -6 \pmod{11}

allows standard integer factorisation.

4

Factor the quadratic polynomial and solve for

x

.

(x - 3)(x + 2) \equiv 0 \pmod{11} \implies x \equiv 3 \text{ or } x \equiv -2 \equiv 9 \pmod{11}

.

Since

11

is prime, a product congruent to

0 \pmod{11}

implies at least one factor is congruent to

0 \pmod{11}

.

5

Select the smallest non-negative integer from the valid solution set

\{3, 9\}

.

x = 3

3

is non-negative and strictly smaller than

9

.

Key Concept

Modular Arithmetic Binary Operations and Quadratic Congruences

Estimated Time:3m 0s

Question 9Question →

What is the value of $(-17 \times 5) \pmod{6}$ expressed in standard non-negative remainder form?

-1

1

3

5

Show answer & explanation

Answer: $5$

Answer

5

Multiplying

-17

by

5

yields

-85

. When

-85

is divided by

6

, the closest multiple of

6

below

-85

is

-90

. The non-negative remainder is

-85 - (-90) = 5

, which belongs to the canonical set

\{0, 1, 2, 3, 4, 5\}

.

Step-by-Step Solution

1

Multiply the numbers inside the parentheses

-17 \times 5 = -85

Perform standard multiplication before applying the modulo operation.

2

Express the negative number in terms of the modulus 6

-85 = 6 \times (-15) + 5

Find the largest multiple of 6 less than or equal to -85, which is -90.

3

Extract the canonical non-negative remainder

5

In standard modular arithmetic, remainders must lie in the range

[0, n-1]

, where

n

is the modulus.

Key Concept

Modular Arithmetic and Non-Negative Remainders

Question 10Question →

Find the value of $r$ such that $-29 \equiv r \pmod{8}$ , where $0 \le r < 8$ .

Show answer & explanation

Answer: 3

Answer

The value of

r

in standard non-negative remainder form is

3

.

Dividing

-29

by

8

yields a quotient of

-4

and a remainder of

3

, because

8 \times (-4) + 3 = -29

. Since

0 \le 3 < 8

,

3

is the standard non-negative remainder.

Step-by-Step Solution

1

Express

-29

in the form

8q + r

, where

q

is an integer and

0 \le r < 8

.

-29 = 8(-4) + 3

To find the standard remainder modulo

8

, the remainder

r

must satisfy

0 \le r < 8

.

2

Identify the remainder value

r

.

r = 3

Since

3

lies within the required range

[0, 7]

, it is the canonical remainder.

Key Concept

Modular Arithmetic and Non-Negative Remainders

Question 11Question →

Determine the least non-negative integer congruent to $68 \pmod{9}$ .

Show answer & explanation

Answer: 5

Answer

The least non-negative integer congruent to

68 \pmod{9}

is 5.

Dividing 68 by 9 gives a quotient of 7 and a remainder of 5. Since

0 \le 5 < 9

, the value 5 is the standard non-negative remainder.

Step-by-Step Solution

1

Divide 68 by the modulus 9 using the division algorithm.

68 = 9 \times 7 + 5

Every integer

a

can be written uniquely as

a = nq + r

where

q

is the quotient and

0 \le r < n

.

2

Extract the non-negative remainder

r

.

r = 5

In modular arithmetic, the least non-negative integer congruent to

a \pmod{n}

is the remainder

r

when

a

is divided by

n

.

Key Concept

Modular Arithmetic Remainder

Question 12Question →

What is the smallest non-negative integer $x$ that satisfies the modular congruence $5x + 7 \equiv 2 \pmod{9}$ ?

8

4

1

7

Show answer & explanation

Answer: $8$

Answer

The smallest non-negative integer

x

is 8.

Subtracting

7

from both sides yields

5x \equiv -5 \pmod{9}

, which reduces to

5x \equiv 4 \pmod{9}

. Multiplying both sides by the modular multiplicative inverse of

5

modulo

9

(which is

2

, since

5 \times 2 = 10 \equiv 1 \pmod{9}

) gives

x \equiv 4 \times 2 = 8 \pmod{9}

. Thus, the smallest non-negative integer value is

8

.

Step-by-Step Solution

1

Isolate the term containing

x

by subtracting 7 from both sides of the congruence.

5x \equiv 2 - 7 \equiv -5 \pmod{9}

Standard algebraic balance operations apply under modular arithmetic.

2

Convert the negative remainder

-5

to its canonical non-negative representative in modulo 9.

-5 + 9 = 4 \implies 5x \equiv 4 \pmod{9}

Modular numbers are conventionally expressed in the non-negative range

[0, n-1]

.

3

Find the modular multiplicative inverse of 5 modulo 9 and multiply both sides by it.

Since

5 \times 2 = 10 \equiv 1 \pmod{9}

, the inverse is 2. Multiplying gives

x \equiv 4 \times 2 = 8 \pmod{9}

.

Multiplying by the modular inverse solves for

x

when

\gcd(5, 9) = 1

.

Key Concept

Linear Modular Congruences and Modular Inverses

Estimated Time:1m 30s

Question 13Question →

If $x$ is the least positive integer satisfying the modular congruence $4x + 9 \equiv 3 \pmod{11}$ , what is the value of $(x^2 - 5x + 2) \pmod{11}$ expressed in standard non-negative remainder form?

2

4

7

9

Show answer & explanation

Answer: 9

Answer

The value of

(x^2 - 5x + 2) \pmod{11}

in standard non-negative remainder form is 9.

Solving

4x + 9 \equiv 3 \pmod{11}

yields

4x \equiv -6 \equiv 5 \pmod{11}

. Multiplying by 3 (the inverse of 4 mod 11) gives

x \equiv 15 \equiv 4 \pmod{11}

. Evaluating

(x^2 - 5x + 2)

at

x = 4

gives

16 - 20 + 2 = -2

. Converting

-2

into the standard non-negative remainder range

[0, 10]

gives

-2 + 11 = 9

.

Step-by-Step Solution

1

Isolate the variable term in the linear modular congruence

4x \equiv 3 - 9 \equiv -6 \pmod{11}

Subtract 9 from both sides of the congruence.

2

Convert the negative right-hand side to a non-negative residue modulo 11

4x \equiv -6 + 11 \equiv 5 \pmod{11}

Add the modulus 11 to obtain the canonical non-negative equivalent.

3

Solve for

x

by multiplying by the multiplicative inverse of 4 modulo 11

Since

4 \times 3 = 12 \equiv 1 \pmod{11}

, multiply both sides by 3:

x \equiv 5 \times 3 = 15 \equiv 4 \pmod{11}

. Thus, the least positive integer is

x = 4

.

The modular inverse of 4 modulo 11 is 3.

4

Substitute

x = 4

into the expression

(x^2 - 5x + 2)

4^2 - 5(4) + 2 = 16 - 20 + 2 = -2

Evaluate the quadratic expression using the calculated value of

x

.

5

Express

-2

in canonical non-negative remainder form modulo 11

-2 + 11 = 9 \pmod{11}

Add the modulus 11 to convert the negative result to a non-negative remainder within

[0, 10]

.

Key Concept

Solving linear modular congruences and converting negative remainders to canonical form

Question 14Question →

What is the canonical non-negative remainder when $3^{45} - 11$ is divided by $7$ ?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.

Using modular exponentiation,

3^3 = 27 \equiv -1 \pmod{7}

, so

3^{45} = (3^3)^{15} \equiv (-1)^{15} = -1 \pmod{7}

. Subtracting 11 yields

-1 - 11 = -12 \pmod{7}

. Adding 14 (a multiple of 7) gives

-12 + 14 = 2

, which is the canonical non-negative remainder.

Step-by-Step Solution

1

Evaluate

3^{45} \pmod{7}

using power rules

3^{45} \equiv -1 \pmod{7}

Since

3^3 = 27 \equiv -1 \pmod{7}

, raising both sides to the power of 15 gives

(3^3)^{15} \equiv (-1)^{15} = -1 \pmod{7}

.

2

Subtract 11 from the modular result

-12 \pmod{7}

Replacing

3^{45}

with

-1

gives

-1 - 11 = -12

.

3

Reduce -12 to its canonical non-negative remainder in

[0, 6]

2

Adding the smallest multiple of 7 that makes the result non-negative gives

-12 + 14 = 2

.

Key Concept

Modular Exponentiation and Negative Remainder Reduction

Question 15Question →

What is the value of $(-43 \times 15) \pmod{8}$ expressed in standard non-negative remainder form?

3

-5

5

-3

Show answer & explanation

Answer: $3$

Answer

3

Reducing

-43

modulo

8

gives

5

, and

15

modulo

8

gives

7

. Multiplying these results yields

35 \equiv 3 \pmod{8}

, which is the canonical non-negative remainder.

Step-by-Step Solution

1

Reduce individual factors modulo 8

-43 \equiv 5 \pmod{8}

(since

-43 = -6 \times 8 + 5

) and

15 \equiv 7 \pmod{8}

(since

15 = 1 \times 8 + 7

)

Simplifying terms before multiplication makes arithmetic easier.

2

Multiply the reduced remainders

5 \times 7 = 35

Modular arithmetic preserves multiplication.

3

Reduce the product modulo 8 to obtain the canonical non-negative remainder

35 = 4 \times 8 + 3 \implies 35 \equiv 3 \pmod{8}

The final answer in modular arithmetic must fall within the range

[0, n-1]

.

Key Concept

Modular Arithmetic and Canonical Non-Negative Remainders

Question 16Question →

If $x$ is the smallest non-negative integer satisfying the modular congruence $7x + 4 \equiv 2 \pmod{13}$ , find the value of $(x^3 + 2x) \pmod{13}$ expressed as a canonical non-negative remainder.

Show answer & explanation

Answer: 6

Answer

The canonical non-negative remainder is 6.

Subtracting 4 from both sides of

7x + 4 \equiv 2 \pmod{13}

gives

7x \equiv -2 \equiv 11 \pmod{13}

. Multiplying by the modular inverse of 7 (which is 2, since

7 \times 2 = 14 \equiv 1 \pmod{13}

) yields

x \equiv 22 \equiv 9 \pmod{13}

. Evaluating

(9^3 + 2 \times 9) \pmod{13}

gives

(729 + 18) = 747

. Dividing 747 by 13 gives a quotient of 57 and a remainder of 6.

Step-by-Step Solution

1

Isolate the linear term in the congruence

7x \equiv -2 \equiv 11 \pmod{13}

Subtracting 4 from both sides simplifies the equation, and

-2 + 13 = 11

converts the negative remainder to positive form.

2

Solve for

x

by multiplying by the multiplicative inverse of 7 modulo 13

x \equiv 9 \pmod{13}

, so

x = 9

Since

7 \times 2 = 14 \equiv 1 \pmod{13}

, multiplying

7x \equiv 11 \pmod{13}

by 2 yields

x \equiv 22 \equiv 9 \pmod{13}

.

3

Evaluate

(x^3 + 2x) \pmod{13}

using modular reduction

6

9^3 = 729 = 56 \times 13 + 1 \equiv 1 \pmod{13}

and

2 \times 9 = 18 = 1 \times 13 + 5 \equiv 5 \pmod{13}

. Adding these gives

1 + 5 = 6

.

Key Concept

Solving linear modular congruences and modular polynomial evaluation

Estimated Time:1m 30s

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